Question

Evaluate the definite integral:

`int_0^(pi/4) (sin x +  cos x)/(9+16sin 2x) dx`

Answer 1

Let `I = int_0^(pi/4) (sin x + cos x)/(9 + 16 sin 2x)`dx

Put sin x - cos x = t 

(cos x + sin x)dx = dt

and 1 - 2 sin x cos x = t²

⇒ sin 2x = 1 - t²

When x = `pi/4`, t = sin `pi/4 - cos  pi/4`

`= 1/sqrt2 - 1/sqrt2 = 0`

When x = 0, t = sin 0 - cos 0 = - 1

`therefore int_0^(pi/4) (sin x + cos x)/(9 + 16 sin 2x)`dx

`= int_(- 1)^0 dt/(9 + 16 (1 - t^2))`

`= int_(- 1)^0 dt/(25 - 16 t^2)`

`= 1/16 int_(- 1)^0 dt/((5/4)^2 - t^2)`

`= 1/16 * 1/(2 * 5/4) [log |(5/4 + t)/(5/4 - t)|]_(-1)^0`

`= 1/40 [log 1 - (log 1 - log 9)]`

`= 1/40 log 9`